Results 1 to 5 of 5

Math Help - find the solution of this PDE !

  1. #1
    Junior Member
    Joined
    Feb 2010
    Posts
    40

    find the solution of this PDE !

    u_{xy}+u_{x} = 0
    subject to boundary conditions :
    u_{y}(0,y)+u(0,y) = ye^{y}
    u(x,0)=x^2

    I have never done direct integration with this " mix " of order, how do i go about it ?!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,707
    Thanks
    1470
    Quote Originally Posted by punkstart View Post
    u_{xy}+u_{x} = 0
    subject to boundary conditions :
    u_{y}(0,y)+u(0,y) = ye^{y}
    u(x,0)=x^2

    I have never done direct integration with this " mix " of order, how do i go about it ?!
    u_{xy}+ u_x= (u_y+ u)_x= 0
    so that u_y+ u is constant with respect to x- that is, u_y+u= f(y) where f is an arbitrary function of y only.

    Now treat u_y+ u= f(y) as an ordinary differential equation in y. It is a first order, linear, differential equation so it can be integrated using an "integrating factor"- that is, a function \mu(y) such that \frac{d\mu u}{dy}= \mu\frac{du}{dy}+ \mu u.

    since \frac{d\mu u}{dy}= \mu \frac{du}{dy}+ u \frac{d\mu}{dy}= \mu\frac{du}{dy}+ \mu u we must have \frac{d\mu}{dy}= \mu or \frac{d\mu}{\mu}= 1, ln(\mu)= y and \mu(y)= e^y.

    That is, e^yu_y+ e^yu= (e^yu)_y= f(y) so that e^y u= F(y)+ C where F(y) is an anti-derivative of f. Since f was arbitrary, F is any arbitrary (differentiable) function of y. However, since we have been doing this with y only, treating x as a constant, that "constant of integration", C, may be an arbitrary function of x-

    e^yu(x,y)= F(y)+ G(x) so u(x,y)= e^{-y}(F(y)+ G(x)) where F and G can be any arbitrary differentiable functions.

    (Yes, just as the solutions to ordinary differential equations involve unknown constants, so the solutions to partial differential equations involve unknown functions).

    Now, use the additional conditions (of which you don't have enough):

     u(x,y)= e^{-y}(F(y)+ G(x)) so u(x, 0)= G(x)= x^2. Now you have u(x,y)= e^{-y}(F(y)+ x^2) but you still need another condition, perhaps something of the form u(0,y)= f(y), to determine F(y).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2010
    Posts
    40

    Solved !

    Thanx HallsofIvy,i used the integrating factor( i don't think i would have thought of that,since never tried to solve a PDE by making it first order!) and with a little manipulation got my answer. Apparently i can't pm yet, but you can verify that the solution is u=\frac{1}{2}e^{y}\left(y-\frac{1}{2}\right)+e^{-y}\left(x^2+\frac{1}{4}\right)
    The key here was that you can rename the product of F(y) and the integrating factor,as G(y) since the product still only depends on y,then the condition on  u_{y}(0,y) + u(0,y) gives the equation G\prime(y)=ye^{2y} it is easy to find the value of G from here, then the last condition makes it easy to find the function in x.
    Last edited by punkstart; April 18th 2010 at 03:22 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    42
    Your answer is correct! Another way would do is something a little different. I'd let v = u_x so the PDE is

    v_y + v = 0 and the second BC u(x,0) = x^2 is u_x(x,0) = 2x so v(x,0) = 2x.

    The PDE is separable so the solution is

    v = F(x)e^{-y} and with the BC gives F(x) = 2x.

    Thus far  v = 2xe^{-y}. <br />

    Then v = u_x = 2xe^{-y} and integrating gives

    u = x^2 e^{-y} + G(y). Now use the first BC

     <br />
u_y + u = G'(y) + G(y) = ye^y \; \text{when}\; x = 0<br />
. This is an ODE for G which can be solved by ODE methods.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2010
    Posts
    40

    Much simpler ! Thank you!

    Will remember that aproach!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Can anybody find the Solution Set of this?
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: February 7th 2011, 04:41 PM
  2. How to find the solution 8*x/e^x = 2
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 26th 2010, 01:38 PM
  3. Replies: 2
    Last Post: September 7th 2009, 02:01 PM
  4. Find the solution
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: August 19th 2009, 11:11 PM
  5. find the general solution when 1 solution is given
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: March 4th 2009, 09:09 PM

Search Tags


/mathhelpforum @mathhelpforum